#calculus 2 triple integral I need help
fading pond
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restive meteor
Let's do this generally.
Suppose we have a sphere of radius R and a plane at distance h < R from the center of the sphere.
It's easier if the sphere is centered at the origin and the plane is parallel to the Oxy plane and above it. What will be the equations of the sphere and the plane?
fading pond
Plane should be paralel
İf the sphere is centered at origin the plane should be z=1
Sphere eq ==> x^2 +y^2+z^2=2^2=4
restive meteor
Well, we are generalizing, so:
x^2 + y^2 + z^2 = R^2
z = h
Alright. Next, let's find where the plane and the sphere intersect.
fading pond
Z=1
x^2 +y^2+z^2=4
x^2 +y^2=3
restive meteor
Yeah. In other words, they intersect at a circle x^2 + y^2 = R^2 - h^2 at height h.
As we are interested in the outer, smaller cap, that region x^2 + y^2 = R^2 - h^2 will be the projection of our cap onto the Oxy plane.
Now, as the cap is above the Oxy plane, it is bounded by two surfaces:
z = h
z = √(R^2 - x^2 - y^2)
So, can you write the integral now in Cartesian coordinates?
fading pond
I tried but I couldnt
restive meteor
Well, the cap is above z = h and below z = √(R^2 - x^2 - y^2), right?
So, what will be the limits for z?
restive meteor
No.
fading pond
Why we are writing h and R insead of their values
restive meteor
Why we are writing h and R insead of their values
Well, it's better to generalize if possible.
fading pond
√(R^2 - x^2 - y^2)<=z<=h
is this the answer you want?
h=1
R=2
√(R^2 - x^2 - y^2)<=z<=h
√(1 - x^2 - y^2)<=z<=1
I don't know what to do next
restive meteor
√(R^2 - x^2 - y^2)<=z<=h
Well, h is on the bottom. So, z changes from h to √(R^2 - x^2 - y^2).
Next, the region on the Oxy plane is x^2 + y^2 = R^2 - h^2. So, what about the limits for y?
fading pond
First part fixed version:
h<=z<=√(R^2 - x^2 - y^2)
1<=z<=√(4 - x^2 - y^2)
Second part
we need to take the projection of this shape
the formula of the projection is x^2+y^2=3
what should I do to find its limits?
restive meteor
Try drawing the region first if you're having trouble.
fading pond
restive meteor
Yeah, looks good.
restive meteor
0<=y<=2pi ?
No. We are working in Cartesian coordinates.
restive meteor
Well, as usual, look at what function is on the bottom and at the top.
fading pond
y=sqrt(3-x^2)
or y = 1
sqrt(3-x^2)<=y<=1
oh wait
-sqrt(3-x^2)<=y<=sqrt(3-x^2)
this is for y
and for x
-sqrt(3)<=x<=sqrt(3)
Now what ?
restive meteor
-sqrt(3-x^2)<=y<=sqrt(3-x^2)
Yes.
restive meteor
and for x
-sqrt(3)<=x<=sqrt(3)
Yes.
Well, now you can write the whole integral in Cartesian coordinates.
fading pond
restive meteor
Great!
Now, let's move onto cylindrical coordinates.
First, the z-limits. Qualitatively, they are the same. But you need to substitute the expressions for x and y into the upper limit.
fading pond
restive meteor
No, the r and θ limits are not that.
restive meteor
is r dz dr d(theta correct?
Yes, that is correct.
restive meteor
what are they? what should I try
Look at our circular region again.
fading pond
I replaced x and y with polar equations
restive meteor
No, that's not how it works.
fading pond
what should I do?
restive meteor
Well, we look at the region.
x^2 + y^2 = R^2 - h^2 turns into r = √(R^2 - h^2).
So, our region is enclosed by r = √(R^2 - h^2).
It is also enclosed from all sides. So, what will be the limits for θ?
restive meteor
No, r can only be positive.
fading pond
okay r = sqrt(3) a constant
what should I do now
I don't know how to look at the limits of theta
x=rcos theta
y = rsin theta
r^2=x^2+y^2
restive meteor
I don't know how to look at the limits of theta
Well, our curce encloses the origin from all directions. That means it covers all angles.
So, the limits for θ are from 0 to 2π.
For r, we look at what polar curves are enclosing our region. If there is only one of them, the lower limit for r is 0.
fading pond
0<=theta<=2pi
z is same but we change x and y with polar transformations
so
1<=z<=sqrt(4-x^2-y^2)
to
1<=z<=sqrt(4-r^2)
r is distance to origin,
I am not sure about how to find this
I belive we are searching r from the projection graph
or we are searching it in the 3d one
according to cap
the r = 0 at the top
and sqrt(3) at the bottom
so its limits are
0<=r<=sqrt(3)
if this is correct, the hardest one is left
I don't know much about spherical integral
restive meteor
Very good!
restive meteor
if this is correct, the hardest one is left
I don't know much about spherical in...
Try reviewing spherical coordinates first, then.
fading pond
I don't know how to start finding that
restive meteor
Well, look for theory in the book of your choice, look at some examples.
restive meteor
Well, the equation of our sphere in spherical coordinates is obviously r = R. What about the plane z = h?
fading pond
shouldn't we need to make tranformation to find an equation for both of them?
I didn't understand what you mean by "spherical coordinats is r= R"
I know these, but don't know where to put them
restive meteor
shouldn't we need to make tranformation to find an equation for both of them?
The equation of the sphere of radius R centered at the origin in spherical coordinates is just r = R, which is visible from how we define r in spherical coordinates.
Of course, if you want to make sure, you can substitute the values and you will see, but that seems unnecessary.
You will need to do that for the plane, though.
fading pond
plane equation is
z=3
also we know
z= ro * cos phi
so
does that makes
ro*cos phi = 3
?
restive meteor
Well, just express r from that.
restive meteor
Yeah, and what is r?
fading pond
r is sqrt(3)
but how is it going to help in here? Where will we put it
fading pond
oh okay. ro is same as r
ro*cos phi = 1
ro =1/cos(phi)
ro= sec(phi)
restive meteor
Yes, very good!
Now, look at how our plane r = h/cos(φ) is closer to the origin than our sphere r = R.
So, what will be the limits for r?
fading pond
1<=r<=2 ?
restive meteor
No.
Well, the upper limit is correct.
The lower limit would be 1 if we had a sphere of radius 1 instead of the plane.
fading pond
oh it start from origin so it is 0<=ro<=2
restive meteor
No, it doesn't start from the origin.
Our region doesn't include it.
Well, we just established that the equation of the plane is r = 1/cos(φ).
So, what will the lower limit be?
fading pond
phi can get between 0 and pi
so cos(phi) should get values between
0 and 1
fading pond
sqrt(3)<=ro<=2 ?
fading pond
then we draw it like this right?
restive meteor
Not really sure what you mean by the arrows.
The inner surface is r = 1/cos(φ), the outer surface is r = 2, so the limits for r are from 1/cos(φ) to 2.
fading pond
alright, now what how to find limits for theta
restive meteor
Well, θ has the same meaning as in cylindrical coordinates.
So, the limits for θ are the same as there.
fading pond
0 to 2 pi then
restive meteor
Yes.
Now, φ. This is slightly trickier.
Let me make a picture.
Here is the axial cross-section of our region.
As it contains the z-axis, the lower limit for φ is 0.
And the upper limit for φ is the angle of the triangle that I've shown on the picture.
So, try finding that angle.
fading pond
restive meteor
Fantastic! So, now just write the integral. Make sure to remember what the volume element looks like in spherical coordinates.
fading pond
restive meteor
Great!
fading pond
thanks for helping
restive meteor
You're welcome!